\(\begin{array}{l}
\begin{array}{|c|c|c|c|}\hline
_P\backslash ^T&20&15&55\\\hline
25&1&2&5\\\hline
30&5&4&2\\\hline
35&8&5&4\\\hline
\end{array}\\
\text{Phương án cơ bản xuất phát:}\\
\begin{array}{|c|c|c|c|}\hline
_P\backslash ^T&20&15&55\\\hline
25&\matrix{1&\\&20}&\matrix{2&\\&5}&\matrix{5&\\&}\\\hline
30&\matrix{5&\\&}&\matrix{4&\\&}&\matrix{2&\\&30}\\\hline
35&\matrix{8&\\&}&\matrix{5&\\&10}&\matrix{4&\\&25}\\\hline
\end{array}\\
\text{Chi phí vận chuyển ban đầu:}\\
f_o = 1.20 + 2.5 + 2.30 + 5.10 + 4.25 = 240\\
\text{Qui 0 cước phí ô chọn}\\
\begin{array}{|c|c|c|c|c|}\hline
_P\backslash ^T&20&15&55&u_i\\\hline
25&\matrix{1&\\&20}&\matrix{2&\\&5}&\matrix{5&\\&}&0\\\hline
30&\matrix{5&\\&}&\matrix{4&\\&}&\matrix{2&\\&30}&1\\\hline
35&\matrix{8&\\&}&\matrix{5&\\&10}&\matrix{4&\\&25}&3\\\hline
v_j&1&2&1\\\hline
\end{array}\\
\Delta_{11} = 0;\quad \Delta_{12} = 0;\quad \Delta_{13} = -4\\
\Delta_{21} = -3;\quad \Delta_{22} = -1;\quad \Delta_{23} = 0\\
\Delta_{31} = -4;\quad \Delta_{32} = 0;\quad \Delta_{33} = 0\\
\Rightarrow \text{Phương án tối ưu}\\
\text{Vậy phương án tối ưu của bài toán là:}\\
X = \left(\matrix{20&5&0\\0&0&30\\ 0&10&25} \right);\ f_{\min} = 240
\end{array}\)
